The freezing point of benzene decreases by 0.45 ^ ° C on adding 0 ....

$\begin{cases}2{CH}_{3}COOH & ⇌ & {\left({CH}_{3}COOH\right)}_{2} \\ 1 - \alpha & & \dfrac{\alpha}{2}\end{cases}$

$i = 1 - \dfrac{\alpha}{2}$

${\Delta T}_{f} = i \times {K}_{f} \times m$

$0 .45 = i \times 5 .12 \times \dfrac{0 .2 \times 1000}{60 \times 20}$

$i = 0 .527 = 1 - \dfrac{\alpha}{2}$

$⇒ \alpha = 0 .946$ or $94 . 6$